How does $b^2=a^2-ab$ lead to $\frac{b}{a}=\frac{-1+\sqrt 5}{2}$? So basically, I came across this particular solution of a problem that starts with

$$\frac {b}{a-b}=\frac ab \quad\implies\quad b^2=a^2-ab \quad\implies\quad\frac ba=\frac{-1+\sqrt 5}{2}$$

Can someone please explain how the $\frac ba=\frac{-1+\sqrt 5}{2}$ part came about?
The $\frac {b}{a-b}=\frac ab$ part is given as part of the problem.
 A: So we start here:
$$b^2 = a^2 - ab$$
(You can get there just by cross-multiplying in your original equation, the one with the fractions.) Move everything to the left-hand side first:
$$b^2 + ab - a^2 = 0$$
Divide both sides by $a^2$:
$$\frac{b^2}{a^2} + \frac{b}{a}  - 1 = 0$$
To make life easier, define the variable $x=b/a$:
$$x^2 + x - 1 =0$$
This is a quadratic equation in $x$, which is easily solved via the quadratic formula:
$$x = \frac{-(1) \pm \sqrt{(1)^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$$
But since $x = b/a$, then we have the desired
$$\frac b a = \frac{-1 \pm \sqrt{5}}{2}$$
(I'm not sure why what you were given neglects the negative solution of $(-1-\sqrt 5)/2$, as far as I know it's valid.)
A: $b^{2}=a^{2}-ab$ can be written as $x^{2}+x-1=0$ where $x=\frac b a $. Solve this quadratic. You get $\frac b a =\frac {-1\pm \sqrt 5} 2$.
A: Dividing $b^2=a^2-ab$ by $a^2$ yields
$$
\left(\frac b a\right)^2 = 1-\frac b a.
$$
Let $z=\tfrac b a$ to get the quadratic equation $z^2=1-z$ or equivalently
$$
z^2+z-1 = 0.
$$
Solve this using your favorite method of solving a quadratic equation to obtain
$$
z = \frac { -1 \pm \sqrt{5} } 2.
$$
A: As you have to find $\dfrac{a}{b}$ let  $a=bt$ and form a quadratic equation in $t$ to get $$ t=\dfrac{1}{t-1}$$
